Introduction to Open Data Science - Course Project

About the project

This will be my (i.e. Baran’s) project page for the assignments of the course PHD-302 Introduction to Open Data Science, University of Helsinki.

Assignment 1:

Here is my GitHub repository: https://github.com/bbayraktaroglu/IODS-project

date()
## [1] "Tue Nov 29 05:23:51 2022"

Some thoughts about the course:

I have never learned R in my previous programming courses, so this was a little bit of an overwhelming start for me. This is also officially my first time using GitHub. I tried to use it before for some personal projects, but failed to understand its general use. I recently heard about the course through the DONASCI emailing list, and I previously heard about the course from a friend of mine who also suggested for me to take the course. I expect to learn whatever I can about data science in an advanced level. The last time I took a rigorous statistics course (or any programming language course) was during my bachelor studies.

##Thoughts about Exercise set 1 and the R for Health Data Science book:

It seems that the exercise set and the first few chapters of the book provide the basic essentials for learning R as a programming language, with its own quirks and conventions. I found the “pipe function”, i.e. “%>%”, the most out of ordinary way of assigning an input to a function. It is sometimes hard to understand why it is used instead of the generic way of computing a function. Other than this, R seems to be an intuitive language, with easy to understand commands.

R Markdown seems to be a very neat notebook like Latex compilers, or Jupyter notebook. I found it easy to understand, but it will take time to get used to its various syntax.


Assignment 2: Linear Regression

This week I have worked on linear regression. To be honest, last time I studied this subject was almost 8 years ago during my bachelor studies, and although the subject is quite easy, I still find some parts quite fascinating. I have never worked with R, so this week was more of an introduction to hands-on R experience compared to last week’s assignment. R syntax seems intuitive, compared to less user-friendly languages like C or even Java.

date()
## [1] "Tue Nov 29 05:23:51 2022"

2.1 Data wrangling

The task for data wrangling seemed daunting at first, but the individual steps were already built from the ground up in the exercise set, so I have not gotten into any trouble.

2.2 Analysis

Setting up the packages

library(tidyverse)
library(dplyr)
library(ggplot2) 
library(GGally)            
library(purrr)

2.2.1: Reading the dataset

# reading the required file for the assignment
students2014 <- read.csv("learning2014.csv", sep = ",", header = TRUE)

We now compute the dimensions of the data and look at its structure:

dim(students2014)
## [1] 166   7
str(students2014)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : chr  "F" "M" "F" "M" ...
##  $ age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: num  3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : int  25 12 24 10 22 21 21 31 24 26 ...

Description of the dataset:

There are 166 observations (each representing a student) and 7 variables in this dataset. The data as a whole was collected as a survey between 2014 and 2015. The variables which are selected for this assignment try to keep track of what type of pedagogical learning method the students used, their overall attitude towards statistics, together with information about their gender, age and exam points. Here is a table with the definitions of the variables:

Variable Variable Type Definition
gender Character gender of the student, M(male)/F(female)
age Integer age of the student
attitude Numeric (double) average of student’s overall attitude toward statistics, scale between 1-5
deep Numeric (double) deep learning metric, scale between 1-5
stra Numeric (double) strategic learning metric, scale between 1-5
surf Numeric (double) surface learning metric, scale between 1-5
points Integer exam points of the student, scale between 1-5

2.2.2: Graphical overview and summaries

We draw a graphical overview of the dataset:

ggpairs(students2014, mapping = aes(col=gender, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))

We also show summaries of the variables:

summary(students2014)
##     gender               age           attitude          deep      
##  Length:166         Min.   :17.00   Min.   :1.400   Min.   :1.583  
##  Class :character   1st Qu.:21.00   1st Qu.:2.600   1st Qu.:3.333  
##  Mode  :character   Median :22.00   Median :3.200   Median :3.667  
##                     Mean   :25.51   Mean   :3.143   Mean   :3.680  
##                     3rd Qu.:27.00   3rd Qu.:3.700   3rd Qu.:4.083  
##                     Max.   :55.00   Max.   :5.000   Max.   :4.917  
##       stra            surf           points     
##  Min.   :1.250   Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.625   1st Qu.:2.417   1st Qu.:19.00  
##  Median :3.188   Median :2.833   Median :23.00  
##  Mean   :3.121   Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.625   3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :5.000   Max.   :4.333   Max.   :33.00

Comments about the outputs

One can see from the graphical overview the scatter plot, the correlations, and the probability distributions of pairs of each of the variables. And from the summary, one can see the various minima, maxima and mean. Female gender is colored in red, while the male gender is colored in blue.

We see that there are considerably more females than males in the study. Females seem to be much younger than the average male, and the females’ attitude towards statistics seem to be considerably lower than their male counterparts. There seems to be a strong positive correlation between attitude and exam points, for both genders. Interestingly, there is a strong negative correlation between attitude and surface learning for males, while there is no significant conclusion for females. Similarly for the correlation between surface and deep learning. Negative correlation means that male students who prefer surface learning are more likely to have a negative attitude towards statistics.

2.2.3: Model fitting

We choose the variables attitude, strategic learning and surface learning as explanatory variables, and construct a linear regression for the dependent variable “exam points”.

my_model <- lm(points ~ attitude + stra +surf, data = students2014)
summary(my_model)
## 
## Call:
## lm(formula = points ~ attitude + stra + surf, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.0171     3.6837   2.991  0.00322 ** 
## attitude      3.3952     0.5741   5.913 1.93e-08 ***
## stra          0.8531     0.5416   1.575  0.11716    
## surf         -0.5861     0.8014  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08

Comments about the result:

One can see the the adjusted R-squared value is 0.1927, which means the variables attitude, strategic learning and surface learning can explain up to 19.27% deviation within the exam points of a student. Moreover, attitude is considerably significant with a p-value of about \(1.93*10^{-8}\), much less than the general lowest threshold of 0.001. Unfortunately, the other variables are not significant, with p-values above 0.1. So it is highly unlikely that strategic learning and surface learning have an explanatory power as much as attitude. The model has an overall p-value of \(3.156*10^{-8}\), which is very low, so the model is significant overall.

Another model

We now remove the variables stra and surf, since both are not very statistically significant, and try to form a new model:

my_model2 <- lm(points ~ attitude , data = students2014)
summary(my_model2)
## 
## Call:
## lm(formula = points ~ attitude, data = students2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  11.6372     1.8303   6.358 1.95e-09 ***
## attitude      3.5255     0.5674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

Comments about the new result:

As expected, the new result improved the statistical significance of the remaining explanatory variable “attitude”, to about \(4.12*10^{-9}\). The overall p-value is also the same as the one for attitude since we are now using a univariate linear regression. Thus, when we compare this model to the previous one, there has been a significant increase in the trustworthiness of the model. But the adjusted R-square value is now 0.1856, which is lower than the previous one. This means that the model has lost some explanatory power, and now can explain up to 18.56% deviation within the exam points. This is expected, since if we remove variables from a model, the explanatory power is expected to decrease, but not by much. The multiple R-squared is not an important metric in this case, since we only have one explanatory variable.

2.2.4: Diagnostics

# place the following four graphics in same plot
par(mfrow = c(2,2))
# draw diagnostic plots for the final model
plot(my_model2, which = c(1,2,5))

Final comments about the diagnostics:

The final model seems to be fitting our expectations. Q-Q plot is mostly along the line, which means that the distribution of the model mostly follows that of the normal distribution. Residuals vs Fitted plot shows us that most of the points follow along the line residual=0 in a horizontal strip, which means that the result is well-behaved. There are no obvious outliers, and the result seems random enough. So the assumption of linearity is well-supported. Finally, Residuals vs Leverage plot tells us that there are two data points (namely 56 and 35) sitting very close to Cook’s distance, but they do not fall outside of it. Thus none of the data points possess any influential effect on the regression model, but further analysis on the data points 56 and 35 can be made just to be sure.


Assignment 3: Logistic Regression

This week I have worked on logistic regression. Slowly but surely, I am starting to feel comfortable with R and RMarkdown. I hope next week is going to be even more easier for me. Using some peer reviews that I have obtained last week, there was an overhaul in my course diary. Now, it should look nicer.

date()
## [1] "Tue Nov 29 05:24:08 2022"

3.1: Data wrangling

This week, data wrangling felt considerably easy. I followed the tasks and used some help from the Exercise 3. The R code of the data wrangling part is in the data folder of my Github repository. I will put the link here as well: https://github.com/bbayraktaroglu/IODS-project/blob/master/data/create_alc.R

3.2 Analysis

Setting up the packages

library(tidyverse)
library(tidyr)
library(dplyr)
library(ggplot2)
library(readr)
library(boot)
library(GGally)            
library(purrr)
library(gmodels)
library(knitr)
library(patchwork)
library(finalfit)
library(stringr)
library(caTools) 
library(caret)

3.2.1: Reading the dataset

# set the working directory
setwd("/Users/barancik/Github/IODS-project/data")
# reading the required file for the assignment
alc <- read.csv("alc.csv", sep = ",", header = TRUE)

We now compute the dimensions of the data and look at its structure:

dim(alc)
## [1] 370  35
str(alc)
## 'data.frame':    370 obs. of  35 variables:
##  $ school    : chr  "GP" "GP" "GP" "GP" ...
##  $ sex       : chr  "F" "F" "F" "F" ...
##  $ age       : int  18 17 15 15 16 16 16 17 15 15 ...
##  $ address   : chr  "U" "U" "U" "U" ...
##  $ famsize   : chr  "GT3" "GT3" "LE3" "GT3" ...
##  $ Pstatus   : chr  "A" "T" "T" "T" ...
##  $ Medu      : int  4 1 1 4 3 4 2 4 3 3 ...
##  $ Fedu      : int  4 1 1 2 3 3 2 4 2 4 ...
##  $ Mjob      : chr  "at_home" "at_home" "at_home" "health" ...
##  $ Fjob      : chr  "teacher" "other" "other" "services" ...
##  $ reason    : chr  "course" "course" "other" "home" ...
##  $ guardian  : chr  "mother" "father" "mother" "mother" ...
##  $ traveltime: int  2 1 1 1 1 1 1 2 1 1 ...
##  $ studytime : int  2 2 2 3 2 2 2 2 2 2 ...
##  $ schoolsup : chr  "yes" "no" "yes" "no" ...
##  $ famsup    : chr  "no" "yes" "no" "yes" ...
##  $ activities: chr  "no" "no" "no" "yes" ...
##  $ nursery   : chr  "yes" "no" "yes" "yes" ...
##  $ higher    : chr  "yes" "yes" "yes" "yes" ...
##  $ internet  : chr  "no" "yes" "yes" "yes" ...
##  $ romantic  : chr  "no" "no" "no" "yes" ...
##  $ famrel    : int  4 5 4 3 4 5 4 4 4 5 ...
##  $ freetime  : int  3 3 3 2 3 4 4 1 2 5 ...
##  $ goout     : int  4 3 2 2 2 2 4 4 2 1 ...
##  $ Dalc      : int  1 1 2 1 1 1 1 1 1 1 ...
##  $ Walc      : int  1 1 3 1 2 2 1 1 1 1 ...
##  $ health    : int  3 3 3 5 5 5 3 1 1 5 ...
##  $ failures  : int  0 0 2 0 0 0 0 0 0 0 ...
##  $ paid      : chr  "no" "no" "yes" "yes" ...
##  $ absences  : int  5 3 8 1 2 8 0 4 0 0 ...
##  $ G1        : int  2 7 10 14 8 14 12 8 16 13 ...
##  $ G2        : int  8 8 10 14 12 14 12 9 17 14 ...
##  $ G3        : int  8 8 11 14 12 14 12 10 18 14 ...
##  $ alc_use   : num  1 1 2.5 1 1.5 1.5 1 1 1 1 ...
##  $ high_use  : logi  FALSE FALSE TRUE FALSE FALSE FALSE ...

Description of the dataset:

There are 370 observations (each representing a student) and 35 variables in this dataset. The data as a whole was collected as a survey on 27.11.2014, from two different Portuguese schools. The data consists of measurements regarding success of the students in two different subjects: Mathematics and Portuguese language. The variables in this assignment try to keep track of some background information about the student, like age, sex, etc., and important measures regarding the success of the students such as number of past class failures, number of school absences, current health status, alcohol consumption, etc. Some of the variables are binary like the sex or internet access, some are numeric, and some are nominal answers like ‘mother’s job’. Some numeric ones are between 1-5, while some are not bounded. The grades (G1, G2, G3) are between 0-20, and each represent grades obtained in different periods.

The dataset for maths and Portuguese language are combined by taking averages, including the grade variables. We combined the data into one single data which only includes the students who took both courses. The variable ‘alc_use’ is the average of workday alcohol consumption and weekend alcohol consumption. ‘high_use’ is TRUE if ‘alc_use’ is higher than 2 and FALSE otherwise.

3.3.2: Hypotheses

Our main aim is to understand the relationship between alcohol consumption and other variables in the data. We choose the variables ‘failures’, ‘absences’, ‘sex’ and ‘famrel’. We hypothesize that there is a correlation between ‘high_use’ and ‘failures’ (number of past failures), ‘absences’ (number of school absences) and ‘famrel’ (quality of family relations). We also hypothesize that there is a correlation between being a male and high consumption of alcohol.

Having high alcohol consumption should in principle be correlated with the number of past failures, since the student might have a serious alcohol problem, thus creating high number of failures.

Similarly, high alcohol consumption is expected to be correlated with high number of absences, since if the student is intoxicated almost always, then attending a class becomes difficult if not impossible.

For family relations, we expect that bad family relations is correlated with high alcohol consumption, since students may try to escape from troublesome relations at home and alcohol is one such solution.

Finally, we expect high alcohol consumption from male students, but we accept that this could be read off as a sexist expectation.

3.3.3 Plots

We now draw some plots regarding high alcohol usage versus the hypothesized variables above:

# put the hypothesized  variables in new data frame
keep_columns <- c("high_use", "failures", "absences", "famrel", "sex")
alc_hypo <- select(alc, one_of(keep_columns))

Let’s now draw a scatter plot to first summarize everything:

ggpairs(alc_hypo, mapping = aes(col=sex, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))

Now let’s start with a bar plot between high alcohol consumption and sex:

# initialize a plot of 'high_use'
g1 <- ggplot(data = alc, aes(x = high_use))

# draw a bar plot of high_use by sex
g1 + geom_bar()+facet_wrap("sex")

We can see that there can definitely be some correlation with being a male and having high alcohol consumption. Percentage of females who drink is very small compared to females who do not. But this ratio increases for males.

We now construct a bar plot of each variable:

# initialize a plot of 'high_use'
g2 <- ggplot(data = alc, aes(x = high_use))

# draw a bar plot of high_use by failures
g2 + geom_bar()+facet_wrap("failures")

We see that eventually, similar to sex, the ratio of high to low alcohol consumption increases as the number of past failures increase. So there could be some correlation.

# initialize a plot of 'high_use'
g3 <- ggplot(data = alc, aes(x = high_use))

# draw a bar plot of high_use by absences
g3 + geom_bar()+facet_wrap("absences")

We see that similar to ‘sex’ and ‘failures’, the ratio of high to low alcohol consumption increases as the number of absences increase. So there could again be some correlation.

# initialize a plot of 'high_use'
g3 <- ggplot(data = alc, aes(x = high_use))

# draw a bar plot of high_use by family relations
g3 + geom_bar()+facet_wrap("famrel")

Finally for family relations, we again have a similar situation, but it is a little bit complicated. Overall, it seems again that the ratio of high to low alcohol consumption increases as the family relations get worse.

We also draw a bar plot which includes all of our explanatory variables, together with the dependent variable:

# draw a bar plot of each variable
gather(alc_hypo) %>% ggplot(aes(value)) + geom_bar()+ facet_wrap("key", scales = "free")

Finally, a boxplot of family relations and absences by alcohol consumption and sex:

# initialize a plot of high_use and family relations
g1 <- ggplot(alc, aes(x = high_use, y = famrel, col = sex))

# define the plot as a boxplot and draw it
g1 + geom_boxplot() + ylab("family relations")+ggtitle("Student family relations by alcohol consumption and sex")

# initialize a plot of high_use and absences
g2<- ggplot(alc, aes(x = high_use, y = absences, col = sex))

# define the plot as a box plot and draw it
g2 + geom_boxplot() + ylab("absences") +ggtitle("Student absences by alcohol consumption and sex")

Overall observations

We see that there could be some correlation between the hypothesized explanatory variables (failures, absences, sex, family relations) and the dependent variable (high alcohol consumption). We will further analyze this.

3.3.4 Using logistic regression

We now move onto a more statistical way of showing why our hypotheses are (significantly) true. We will use logistic regression to accomplish this:

# find the model with glm()
m <- glm(high_use ~ failures + absences + sex + famrel, data = alc, family = "binomial")

# print out a summary of the model
summary(m)
## 
## Call:
## glm(formula = high_use ~ failures + absences + sex + famrel, 
##     family = "binomial", data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.0786  -0.8216  -0.5746   0.9760   2.1820  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -0.79406    0.54281  -1.463  0.14350    
## failures     0.57328    0.20531   2.792  0.00523 ** 
## absences     0.08941    0.02274   3.932 8.43e-05 ***
## sexM         1.04800    0.25091   4.177 2.96e-05 ***
## famrel      -0.29791    0.13044  -2.284  0.02238 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 401.77  on 365  degrees of freedom
## AIC: 411.77
## 
## Number of Fisher Scoring iterations: 4
# print out the coefficients of the model
coef(m)
## (Intercept)    failures    absences        sexM      famrel 
## -0.79406437  0.57327802  0.08940969  1.04800182 -0.29791173
# compute odds ratios (OR)
OR <- coef(m) %>% exp

# compute confidence intervals (CI)
CI<- exp(confint(m))
## Waiting for profiling to be done...
# print out the odds ratios with their confidence intervals
cbind(OR, CI)
##                    OR     2.5 %    97.5 %
## (Intercept) 0.4520039 0.1532656 1.2982302
## failures    1.7740730 1.1936470 2.6881229
## absences    1.0935286 1.0480739 1.1462668
## sexM        2.8519467 1.7556470 4.7047758
## famrel      0.7423669 0.5735490 0.9583804

We now interpret the summary. Observe that absences and sex (male) have a highly significant (positive, since coefficient is positive) correlation with high_use, with p-value less than 0.001. Failures have a significant (positive) correlation with high_use, with p-value between 0.01 and 0.001. Finally, family relations have a significant (negative, since the coefficient is negative) correlation with high_use, with p-value between 0.05 and 0.01. All of our hypotheses can be accepted and are indeed significant enough. If one surmises that the p-value should be less than 0.01 to achieve even greater significance, then family relations loses its significant correlation with high-use.

We now interpret the coefficients as odd ratios. Note that the exponentials of the coefficients of a logistic regression model can be interpreted as odds ratios between a unit change (vs. no change) in the corresponding explanatory variable:

  1. We see that odd ratio of failure is about 1.77. This means that for each unit of failure, the increase in odds of having a student with high alcohol consumption is about 1.77 times. Thus, more failures mean higher odds of having high alcohol consumption, as hypothesized earlier.

  2. Similarly, for each unit of absences, the increase in odds of having a student with high alcohol consumption is about 1.09 times, which is very close to 1, thus there is almost no change in high alcohol consumption, but it is still greater than 1, so it is in line with our hypothesis.

  3. Odds ratio for sex (male) is about 2,85, which indicates that changing sex (i.e. increasing the odds of being a male), alters the odds of having a student with high alcohol consumption the most. This is also in line with our hypothesis, since we said that being a male should be positively correlated with high alcohol consumption.

  4. Finally, odds ratio of family relations is less than 1, which means that we are losing in the odds of having high alcohol consumption if we increase family relations. This also is in line with our hypothesis: better family relations=low alcohol consumption.

3.3.5 Predictive power of the model

We compute the predictive power of the model with failures, absences, sex and family relations as our explanatory variables and high_use as the dependent variable. We excluded none of the initial choice for the explanatory variables, since in the last section we found a significant correlation between them and high_use.

# fit the model
m <- glm(high_use ~ failures + absences + sex + famrel, data = alc, family = "binomial")

# predict() the probability of high_use
probabilities <- predict(m, type = "response")

library(dplyr)
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)

# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability>0.5)

# see the last ten original classes, predicted probabilities, and class predictions
select(alc, failures, absences, sex, famrel, high_use, probability, prediction) %>% tail(10)
##     failures absences sex famrel high_use probability prediction
## 361        0        3   M      4    FALSE   0.3386132      FALSE
## 362        1        0   M      4    FALSE   0.4098873      FALSE
## 363        1        7   M      5     TRUE   0.4908822      FALSE
## 364        0        1   F      5    FALSE   0.1002713      FALSE
## 365        0        6   F      4    FALSE   0.1901165      FALSE
## 366        1        2   F      5    FALSE   0.1777706      FALSE
## 367        0        2   F      4    FALSE   0.1410142      FALSE
## 368        0        3   F      1    FALSE   0.3049689      FALSE
## 369        0        4   M      2     TRUE   0.5039381       TRUE
## 370        0        2   M      4     TRUE   0.3188873      FALSE
# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction)
##         prediction
## high_use FALSE TRUE
##    FALSE   244   15
##    TRUE     77   34

We see that our model correctly predicts 244 false and 34 true observations. The rest are inaccurately classified individuals. We can graph the actual values vs predictions:

# initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(alc, aes(x = probability, y = high_use),aes(col=prediction))

# define the geom as points and draw the plot
g + geom_point()

# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>%addmargins()
##         prediction
## high_use      FALSE       TRUE        Sum
##    FALSE 0.65945946 0.04054054 0.70000000
##    TRUE  0.20810811 0.09189189 0.30000000
##    Sum   0.86756757 0.13243243 1.00000000

We now compute the average number of inaccurately classified individuals:

# Work with the exercise in this chunk, step-by-step. Fix the R code!
# the logistic regression model m and dataset alc with predictions are available

# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}

# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2486486

We find a number of about 0.25. This means that on average, 1 out of 4 people are inaccurately classified, meaning that they are falsely accused of heavy drinking while actually being light drinkers, or vice versa.

3.3.6: Bonus

We perform 10-fold cross validation:

# K-fold cross-validation
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)

# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2459459

We obtain a number of about 0.26, which is the same if not a little bit worse than the predictions in the Exercise. Thus the test performance is almost identical. This is largely due to family relations having a small impact on the dependent variable, compared to sex or failures, thus including family relations did not create a better model, and may in fact worsen it.


Assignment 4: Clustering and classification

This week I have worked on clustering and classification. This week definitely felt much more easier for me.

date()
## [1] "Tue Nov 29 05:24:30 2022"

4.1: Data wrangling

This week, data wrangling felt even easier than the last week. I mostly used some help from create_alc.R. The R code of the data wrangling part is in the data folder of my Github repository. I will put the link here as well: https://github.com/bbayraktaroglu/IODS-project/blob/master/data/create_human.R

4.2 Analysis

Setting up the packages

library(MASS)
library(dplyr)
library(tidyr)
library(tidyverse)
library(corrplot)
library(ggplot2)
library(plotly)

4.2.1: Reading the dataset

# reading the required file for the assignment
data("Boston")
# checking out its dimension, structure and summary
dim(Boston)
## [1] 506  14
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

In the Boston dataset, there are 506 observations and 14 variables. It is included in the MASS package of R. This data frame contains gathered data related to housing values in suburbs of Boston. Most of the variables are numeric (float), while “chas” and “rad” variables are integers.

4.2.2 Plots

Let’s put our newly learned knowledge about correlation plots to good use. The following is the correlation matrix and its various plots of the Boston data:

# calculating the correlation matrix, also round it to 2 digits
cor_matrix <- cor(Boston) %>% round(digits=2)

# print the correlation matrix
print(cor_matrix)
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00
# visualize the correlation matrix
library(corrplot)
corrplot(cor_matrix, method="circle")

corrplot(cor_matrix, method="number")

Observe that most of the variables are more or less correlated with each other, but the “chas” variable is mostly correlated with itself, while having correlation very close to 0 with the other variables. We know from basic probability theory that uncorrelated data does not imply independence, so we cannot infer that “chas” is independent from the other variables. We can only say that it is almost uncorrelated from the other variables. “rad” and “indus” has high overall positive correlation with most of the other variables (except “chas”). “rad” has 0.91 correlation with “tax” and 0.72 with “indus”. “indus” has -0.71 correlation (strong negative correlation) with “dis”, while “nox” has -0.77 correlation with “dis”.

4.2.3 Standardize the dataset and print scaled data summary

We will scale the data by subtract the column means from the corresponding columns and divide the difference with standard deviation. This normalizes the variables to be centered with standard deviation 1.

# scaling the Boston
boston_scaled <- as.data.frame(scale(Boston))

# summaries of the scaled variables
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

We now create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). We will use the quantiles as the break points in the categorical variable.

We will then drop the old crime rate variable from the dataset. Afterwards, we divide the dataset to train and test sets, so that 80% of the data belongs to the train set.

# creating a categorical variable called "crime" from scaled crime rate
boston_scaled$crim <- as.numeric(boston_scaled$crim)
crime <- cut(boston_scaled$crim, breaks = quantile(boston_scaled$crim), include.lowest = TRUE, label=c("low", "med_low", "med_high", "high"))

# remove original crim from the dataset
boston_scaled <- boston_scaled %>% dplyr::select(-crim)

# add the new categorical variable to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

# number of rows in the Boston dataset
n <- nrow(boston_scaled) 

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# creating the train set
train <- boston_scaled[ind,]

# creating the test set 
test <- boston_scaled[-ind,]

4.2.4 Fit the LDA and draw its (bi)plot

We will now fit the linear discriminant analysis on the train set. We will use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. We then draw the LDA (bi)plot.

# linear discriminant analysis
lda.fit <- lda(crime ~ . , data = train)

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results
plot(lda.fit, dimen = 2, col=classes, pch=classes)
lda.arrows(lda.fit, myscale = 1)

Observe that, the LDA plot predicts “rad” has the most variation in the dataset, towards the mostly “high” cluster.

4.2.5 LDA prediction

set.seed(123)

# saving the correct classes from test data
correct_classes <-test$crime

# removing the crime variable from test data
test <- dplyr::select(test, -crime)

# predicting classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulating the results
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       25      12        1    0
##   med_low    2      13        8    0
##   med_high   0      10        7    1
##   high       0       0        0   23

We find that almost all of the results are accurately predicted. Correctly classified observations are about 67, while the rest (about 35) are incorrectly classified. The inaccuracy rate of the LDA is about 34% (can be as low as 23% in some other sampling with another other seed).

4.2.6 K-means clustering

We reload Boston, rescale it and compute its Euclidean distance.

# reload the data
data("Boston")

# scale the data again
boston_scaled <- as.data.frame(scale(Boston))

# compute the Euclidean distance of Boston
dist_eu <- dist(boston_scaled)

# summary of dist_eu 
summary(dist_eu)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

We now run the k-means algorithm:

set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
## Warning: `qplot()` was deprecated in ggplot2 3.4.0.

The above plot extensively shows us that there is a significant drop at the value 2. Thus, the optimal number of clusters is 2.

We now run k-means algorithm again, this time with 2 clusters, and plot the Boston dataset with the clusters. The clusters will be colored in red and black.

set.seed(123)

# k-means clustering with 2 clusters
km <- kmeans(Boston, centers = 2)

# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

If one zooms in to the plot above, one would see that “rad” has nicely separated clusters across all of the possible pairings. “tax” also has good separation of clusters. The other variables are a complete mess, and no other conclusion can be drawn.

4.2.7 Bonus: K-means algorithm and LDA

We will now perform k-means algorithm on the original Boston data (after scaling). We choose 5 clusters. We then perform LDA using the clusters as target classes. We will include all the variables in the Boston data in the LDA model.

set.seed(5)
# reload the data
data("Boston")

# scale the data again
boston_scaled <- as.data.frame(scale(Boston))

# k-means clustering with 5 clusters
km <- kmeans(Boston, centers = 5)

# linear discriminant analysis on the clusters, with data=boston_scaled, and target variable km$cluster
lda.fit <- lda(km$cluster ~ ., data = boston_scaled)

# target classes as numeric
classes <- as.numeric(km$cluster)

# plot the lda results. Note that lda.arrows is the same function we have used above
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)

Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influential linear separators for the clusters?

We observe in the above biplot that “tax” and “rad” have the most variation in the dataset. Moreover, the K-means seems to form accurate and separate clusters.

4.2.8 Super-Bonus: A cool 3D plot of LDA and K-means

We will recall the code for the (scaled) train data that we used to fit the LDA. We then create a matrix product, which is a projection of the data points.

set.seed(123)
# LDA

lda.fit <- lda(crime ~ ., data = train)

model_predictors <- dplyr::select(train, -crime)

# check the dimensions
dim(model_predictors)
## [1] 404  13
dim(lda.fit$scaling)
## [1] 13  3
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

We now create a 3D plot of the columns of the matrix product:

library(plotly)

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color=~train$crime)

Now let’s run the k-means algorithm on the matrix product with 4 clusters (since the number of clusters of crime is 4), and draw another 3D plot where the color is defined by the clusters of the k-means.

set.seed(5)
km = kmeans(model_predictors, centers = 4)

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color=~factor(km$cluster))

The k-means clustering is mostly successful. One can see that there are 2 superclusters, while the clusters 1,2,4 (mostly) form their own subclusters under one of the superclusters. The cluster 3 is shared between the huge clusters. In the clusters for “crime”, “med_high” has this same property, while the other clusters are nicely separated into two superclusters. Thus, the k-means clustering plot with 4 clusters seems to give similar results compared to the lda.fit of the “crime” variable.


(more chapters to be added similarly as we proceed with the course!)